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DENIUS [597]
3 years ago
5

How many moles are in 6.3x1054 molecules of Ca(C2H202)2?

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
8 0

In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. It is the mass of 1 mole of the substance or 6.022×10²³ particles, expressed in grams.

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Which of the following compounds has the highest boiling point?1. F22. NaF3. HF4. ClF
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HF

Explanation:

Hf has hydrogen bonding which is the strongest intermolecular forces. The stronger the IM forces, the higher the boiling point.

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How does the absorption and release of energy affect temperature change during a chemical reaction?  PLEASE HELP!!!!!!!!!!!!!!!
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Water molecules are polar because the:
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<u>Answer;</u>

B) water molecule has a bent shape.

<u>Explanation;</u>

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3 0
4 years ago
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A chemist wishing to do an experiment requiring 47Ca2+(half-life = 4.536 days) needs 5.00 g of the nuclide. What mass of 47CaCO
Artyom0805 [142]

Answer: The answer is 6.78 grams.

Explanation: The equation used for solving this type of problems  is:

\frac{N}{N_0}=(\frac{1}{2})^n

where, N_0 is the initial amount of radioactive substance, N is the remaining amount and n is the number of half lives.

Number of half lives is calculated on dividing the given time by the half life.

n = time/half life

Time is given as 48.0 hours and the half life is given as 4.536 days. let's make the units same and for this let's convert the half life from days to hours.

4.536days(\frac{24hours}{1day})

= 108.864 hours

So, n=\frac{48.0}{108.864}  = 0.441

Since 5.00 g is the required amount when the radioactive substance is delivered to the scientist, it would be the final amount that is N. We need to calculate the initial amount. Let's plug in the values in the equation:

\frac{5.00}{N_0}=(\frac{1}{2})^0^.^4^4^1

\frac{5.00g}{N_0}=0.737

N_0=\frac{5.00g}{0.737}

N_0 = 6.78 g

So, 6.78 g of the radioactive substance needs to be ordered.

4 0
4 years ago
You mix 265.0 mL of 1.20 M lead(II) nitrate with 293 mL of 1.55 M potassium iodide. The lead(II) iodide is insoluble. What amoun
slava [35]

Answer:

105 grams PbI₂

Explanation:

Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)

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moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant

moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂

grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)

7 0
3 years ago
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